Ground-state hydrogen in wave mechanics – most probable electron–nucleus separation In the quantum (wave-mechanical) description of hydrogen, the radial probability density for the ground state (1s) is maximum at a separation equal to which Bohr radius?

Introduction / Context:Bohr’s model introduced the Bohr radius a0 as the characteristic scale of the hydrogen atom. Quantum mechanics refines this picture with wavefunctions and probability densities. For the 1s state, the most probable electron–proton separation emerges from the radial probability distribution, connecting the old model with modern quantum theory.

Given Data / Assumptions:

• Hydrogen atom, ground state (1s) wavefunction.
• Radial probability density P(r) relates to the likelihood of finding the electron at distance r from the nucleus.
• Bohr radius a0 is the natural length scale.

Concept / Approach:

The ground-state hydrogenic wavefunction yields a radial probability density that peaks at r = a0. Thus, while the expectation values differ (e.g., = 1.5 a0), the most probable radius equals the first Bohr radius. This result reconciles Bohr’s semiclassical orbit radius with the probabilistic quantum description for the ground state.

Step-by-Step Solution:

Verification / Alternative check:

Plotting P(r) vs r shows a clear peak at r = a0; tables of hydrogen expectation values confirm the distinction between mode (a0) and mean (1.5 a0).

Why Other Options Are Wrong:

Second or third Bohr radii correspond to excited states; 'either' is incorrect because the ground state has a single maximum. Zero separation is not the most probable due to the r^2 weighting in the radial probability.

Common Pitfalls:

Confusing the expectation value with the most probable value; misapplying Bohr’s integer-n radii to the 1s quantum distribution without considering probability density.

Final Answer:

first Bohr radius